Problem: In triangle $BCD$, $\angle C = 90^\circ$, $CD = 3$, and $BD = \sqrt{13}$. What is $\tan B$?
Solution: [asy]
pair B,C,D;
B = (0,0);
C = (2,0);
D = (2,-3);
draw(B--C--D--B);
draw(rightanglemark(B,C,D,7));
label("$D$",D,SE);
label("$B$",B,NW);
label("$C$",C,NE);
label("$3$",(C+D)/2,E);
label("$\sqrt{13}$",(B+D)/2,SW);
[/asy]

Because $\triangle BCD$ is a right triangle, we know that $\tan B = \frac{CD}{BC}$.

By the Pythagorean Theorem, $BC = \sqrt{BD^2 - CD^2} = \sqrt{13 - 9} = \sqrt{4} = 2$.

Then $\tan B = \frac{CD}{BC} = \boxed{\frac{3}{2}}$.